Just to be clear, $\epsilon$ and $\epsilon_0$ are the solvent and vacuum dielectric constants, $kT$ is the thermal energy, $z$ is the ion valency and q is the proton charge. Finally, $C_0$ is a reference concentration at which $\psi = 0$.

This is supposed to be a fundamental result for this system but I can't find any explanation to this result or derivation. I can find the solution for a system with excess salt thought.

So, how do I solve this system/where is the flow in my solution?

Edit:
The review I am referring to is David Andelman's chapter in the book structure and dynamics of membranes. The chapter title is: Electrostatic properties of membranes: The Poisson-Boltzmann equation

You seem to have mixed your $x$s' and $z$s'. You say "..a boundary condition at $z=0$.." and then you say "...$z$ is the ion valency...". Can you clean up? Also can you please point to the review...?
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Vijay MurthyMay 1 '12 at 18:33

@VijayMurthy: Thank you for your remark. Nevertheless, the solution I have posted, and my approach are derived under the right notations.
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YotamMay 2 '12 at 6:49

1

It still seems messed up..! $x$ seems to have disappeared from the solution $\psi$. Sorry. But could you please cite the review?
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Vijay MurthyMay 2 '12 at 8:53

OK. I give up..! The notation is not clear. $z$ is sometimes used as the ion valency and sometimes used as a space variable. And no reference to the review even after two requests.
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Vijay MurthyMay 2 '12 at 10:25

@VijayMurthy Sorry I didn't post the source of this. I wasn't in my office (where I have the book) until now. Reading my question I have noticed that I have switched my notepad while typing in. Sorry
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YotamMay 2 '12 at 10:57

1 Answer
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We want to solve the Poisson-Boltzmann (PB) equation for the case of a uniformly charged flat membrane of surface charge density $\sigma$ located at $z=0$. Also we want to consider the case of monovalent ions in the solution with no added electrolyte. We thus want to solve
$$\psi''(z) = -\frac{4\pi e n_0}{\varepsilon_w} e^{-e\psi(z)/T} = -\alpha e^{-\beta \psi(z)}$$
in the half-space $z>0$ where $\alpha= 4\pi e n_0/\varepsilon_w$ and $\beta=e/T$ with the boundary condition
$$\psi'(z) \big\vert_{z=0} = - \frac{4\pi}{\varepsilon_w} \sigma \,\, > 0.$$
Notice that this boundary condition implicitly assumes that the surface charge density is negative $\sigma<0$. This also means that the potential $\psi(z)$ will be an increasing function of $z$.

To solve non-linear equations of the type
$$\psi''(z) = -\alpha e^{-\beta \psi(z)}$$
there is a general trick as outlined here. Using this trick, we arrive at
$$\psi'(z) = \pm \sqrt{\frac{2\alpha}{\beta} e^{-\beta\psi(z)} + C}.$$

Now for some physical input. Since the potential increases with $z$, and we require that the electric field $\psi'(z)$ must vanish for $z\to\infty$, we get $C=0$. This also indicates that we must take the positive root for $\psi'(z)$ above. With this, we get
$$\psi'(z) = \sqrt{\frac{2\alpha}{\beta}} e^{-\beta\psi(z)/2}$$
which is easily solved to give
$$\psi(z) = \frac{2}{\beta} \ln(z+b) + \psi_0 = \frac{2T}{e} \ln(z+b) + \psi_0$$
where
$$\psi_0 = \frac{1}{\beta} \ln \frac{2\alpha}{\beta} = \frac{T}{e} \ln \frac{8 \pi T n_0}{\varepsilon_w}.$$
By imposing the boundary condition, you will easily find the Gouy–Chapman length
$$b=\frac{\varepsilon_w T}{2\pi |\sigma| e}.$$